Sequence converging to different limits with respect to two different _complete_ norms

Do there exist a real vector space $$X$$ complete with respect to norms $$|\cdot|$$ and $$\|\cdot\|$$ and a sequence $$(x_n)_{n\in \mathbb N} \subset X$$ such that there exist $$x,y\in X$$: $$x\ne y$$, $$|x_n - x|\to 0$$ and $$\|x_n -y\|\to 0$$ as $$n\to \infty$$?

Without requiring $$X$$ to be complete with respect to $$|\cdot|$$ and $$\|\cdot\|$$ this is possible. In particular one can repeat Bill Johnson's construction: let $$(e_n)_{n\in\mathbb N}$$ be the standard orthonormal basis of $$\ell_2$$. Let $$A\colon \ell_2 \to \ell_2$$ be a linear operator such that $$A(e_1 + \frac{1}{n} e_n) = \frac{2}{n}e_n$$ for any $$n\in \mathbb N$$ (it can be constructed by extending the linearly independent sets $$(e_1 + \frac{1}{n} e_n)_n$$ and $$(\frac{2}{n} e_n)_n$$ to Hamel bases of $$\ell_2$$ and extending the bijection between these bases to a linear isomorphism). Let $$\|\cdot\|$$ be the standard norm of $$\ell_2$$ and let $$|x| := \|Ax\|$$, $$x\in \ell_2$$. Then clearly $$x_n = e_1 + \frac{1}{n} e_n$$ is the desired sequence (with $$x=0$$ and $$y=e_1$$).

In this example clearly $$(X, \|\cdot\|)$$ is Banach, but it is not clear whether $$(X, |\cdot|)$$ is Banach, as Christopher A. Wong commented in the discussion of a similar question on MSE.

$$|\cdot|$$ is complete. Indeed, suppose $$(x_n)$$ is a Cauchy sequence in $$(X,|\cdot|)$$. Then $$(Ax_n)$$ is a Cauchy sequence in $$(X, \|\cdot\|)$$ which is complete and hence has a limit $$y$$. Since $$A$$ was constructed to be a linear isomorphism, $$y=Ax$$ for some $$x\in X$$. Therefore $$|x_n-x|=\|A(x_n-x)\|=\|Ax_n-y\|$$ tends to $$0$$, proving $$(x_n)$$ is convergent.